AshehadAl wrote: Thanks Stuart for the response, I have followed it to degree. As I did not fully understand, I am attaching two versions of the same program that you wrote with some more explanations. However, I am focusing for simplicity reasons on 3 vectors. Thanks, Ashehad |
Hi Ashehad,
Unfortunately, I can't open the 'mcd' version even in M15, so I don't know what your questions were.
If you want to calculate each combination of vector, say a and b, then you can simplify matters considerably (I think!). If condA(a,b) is true, then condB(b,a) will be true and vice versa. Consequently, the total meeting A and B will be the same. In which case, you need simply count those meeting Condition C and split the difference between this number and the total number of possible combinations to get the number meeting A and B.
Stuart
Note: there was at least one error in condA - a '>' should have been a '<'.